1. Transition Metals

2. Aqueous oxoanions of transition elements.
Figure 23.5
Aqueous oxoanions of transition elements.
Mn2+
MnO42−
MnO4−
One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation states.
VO43−
Cr2O72−
MnO4− 2

3. 3

4. Colors of representative compounds of the Period 4 transition metals.
Figure 23.6
Colors of representative compounds of the Period 4 transition metals.
sodium chromate
nickel(II) nitrate hexahydrate
potassium ferricyanide
zinc sulfate heptahydrate
Titanium(IV) oxide
scandium oxide
manganese(II) chloride tetrahydrate
copper(II) sulfate pentahydrate
vanadyl sulfate dihydrate
cobalt(II) chloride hexahydrate 4

5. 5

6. 6

7. 7

8. Calculating the Concentration of a Complex Ion
Sample Problem 19.13
Calculating the Concentration of a Complex Ion
PROBLEM:
An industrial chemist converts Zn(H2O)42+ to the more stable Zn(NH3)42+ by mixing 50.0 L of M Zn(H2O)42+ and 25.0 L of 0.15 M NH3. What is the final [Zn(H2O)42+]?
Kf of Zn(NH3)42+ is 7.8x108.
PLAN:
Write the reaction equation and Kf expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH3 and therefore it will drive the reaction to completion.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l)
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
[Zn(H2O)42+]initial = (50.0 L)( M)
75.0 L
= 1.3x10-3 M
[NH3]initial =
(25.0 L)(0.15 M)
75.0 L
= 5.0x10-2 M 8

9. Calculating the Concentration of a Complex Ion
Sample Problem 19.13
Calculating the Concentration of a Complex Ion
Since we assume that all of the Zn(H2O)42+ has reacted, it would use 4 times its amount in NH3.
[NH3]used = 4(1.3x10-3 M) = 5.2x10-3 M
[Zn(H2O)42+]remaining = x(a very small amount)
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)42+(aq) + 4H2O(l)
Concentration (M)
Initial
1.3x10-3
5.0x10-2 -
Change
~(-1.3x10-3)
~(-5.2x10-3)
~(+1.3x10-3) -
Equilibrium x
4.5x10-2
1.3x10-3 -
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
= 7.8x108 =
(1.3x10-3)
x(4.5x10-2)4
x = 4.1x10-7 M 9

10. Calculating the Effect of Complex-Ion Formation on Solubility
Sample Problem 19.14
Calculating the Effect of Complex-Ion Formation on Solubility
PROBLEM:
In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0 M hypo. Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13.
PLAN:
Write equations for the reactions involved. Use Ksp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent.
SOLUTION:
AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13
(a)
S = [AgBr]dissolved = [Ag+] = [Br-]
Ksp = S2 = 5.0x10-13 ; S = 7.1x10-7 M
(b)
AgBr(s) Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq)
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq) 10

11. Calculating the Effect of Complex-Ion Formation on Solubility
Sample Problem 19.14
Calculating the Effect of Complex-Ion Formation on Solubility
[Br-][Ag(S2O3)23-]
[S2O32-]2
Koverall = Ksp x Kf =
= (5.0x10-13)(4.7x1013)
= 24
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)23-(aq)
Concentration (M)
Initial -
1.0
Change -
-2S +S +S
Equilibrium - S S S
Koverall = S2
( S)2 S S
= (24)1/2
= 24;
S = [Ag(S2O3)23-] = 0.45 M 11